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Transformer question

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Njguy

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Can a transformers primary coil be overpowered by the secondary coil? This is hypothetical. Lets same your primary coil has the same amount of windings as the secondary coil. So if this were a normal transformer the voltage and current would be the same minus some inefficiency losses. But what if your secondary coil was made of some super special material like carbon-nanotube copper wire (this actually exists in labs). So now your voltage is the same, but you can generate way more current due to the wires radically reduced resistance. Would the secondary coil now be overpowering the primary thus nullifying any gained advantage?
 
No matter what your secondary winding is made of it will not overcome the inefficiency in the coupling between primary and secondary windings, at least in conventional transformers. Also, the primary winding has to be able to actually supply the current required by the secondary, plus losses, without saturating the transformer core, otherwise the secondary voltage will drop in order to supply the current you are drawing. So no, making the secondary winding out of unobtainium will not overpower the primary winding, unless you pull more current with your load than the primary winding can supply nor for that matter, will it overcome the losses in the transformer core :)
 
Hi,

Very clever new word here tunedwolf, "unobtainium", which implies a material that can not be obtained. That's neat.

One of the biggest losses in transformers is the copper losses. To get more current you need thicker wire, and thicker wire leads to a bigger window and a bigger window leads to a bigger transformer. With a given copper wire size and given current rating we end up with a certain amount of heat, and that heat has to be dissipated, and also represents a loss of efficiency.
With a material that could conduct better, we'd see an increase in efficiency. Since the efficiency is so closely related to the copper losses, we should see a marked increase in efficiency.
There are other losses however that would not go away. For example hysteresis loss. We'd still have that to deal with.
So in the end we would end up with a better transformer (smaller, less weight) but we'd also have to improve the core material if we wanted to go any farther.
 
No matter what your secondary winding is made of it will not overcome the inefficiency in the coupling between primary and secondary windings, at least in conventional transformers. Also, the primary winding has to be able to actually supply the current required by the secondary, plus losses, without saturating the transformer core, otherwise the secondary voltage will drop in order to supply the current you are drawing. So no, making the secondary winding out of unobtainium will not overpower the primary winding, unless you pull more current with your load than the primary winding can supply nor for that matter, will it overcome the losses in the transformer core :)

I actually made a mistake in my initial question. I also meant that the secondary would have more wingdings since it is made of 'unobtainium' lol.
 
With a material that could conduct better, we'd see an increase in efficiency. Since the efficiency is so closely related to the copper losses, we should see a marked increase in efficiency.

Not to mention that transformers are one of the most efficient 'machines' there are, so the 'marked' increase in efficiency wouldn't be that high anyway - using silver windings would increase efficiency anyway, but obviously isn't cost effective - using something even more expensive for similar small improvements would be even less worthwhile.
 
Hi Nigel,


Yes that's a good point. If the winding material is too expensive it's going to get harder to justify using it, except of course in space where every little weight decrease is a huge blessing.

Let me be more specific too. The winding resistance plays a big part in the heat generated by the construction, and the heat is of course I^2*R losses for the most part. In the design of the transformer the wire diameter thus plays a big part in the design. We cant design something that puts out 10 amps if the wire diameter normally works at only 5 amps. Doing so would cause a higher P/SA (power over surface area) ratio which would mean of course the transformer works ok at 5 amps but not at 10 amps where it overheats badly. But if we could reduce the resistance by 1/2, we might be able to stretch it to 9 or 10 amps. Thus we'd start with a transformer rated for 100 watts and end up with a transformer of the exact same size that is rated for 200 watts.

I ran into this problem when i purchased a 24vac step down transformer that was supposedly rated for 10 amps out. It turned out that running at around 8 or 9 amps caused it to get so hot that i could not trust it. If i could simply replace the windings with wire that has 1/2 the resistance, this problem would be much lessened and i would bet it would handle the 10 amps.

But yes the cost is an issue too. That takes lots of time to get around. For the first million pieces out the door we'd be paying top dollar. Later as the industry realizes the earning potential they will start to develop cheaper ways to do it, at least hopefully, unless it involves rare earth's and then we are all beat :)

We might be lucky to see this in our lifetime. We read about new stuff all the time but the time it takes to get to us at home is sometimes just a little too long.

[LATER]
After thinking about this for a little longer i'd like to go back to the beginning and see what effect a higher primary current would have because we would have the same inductance in the primary. With the same inductance we'd have higher flux density which could be too much for the core, so we might also need a better core material as well.
 
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.................................Also, the primary winding has to be able to actually supply the current required by the secondary, plus losses, without saturating the transformer core, otherwise the secondary voltage will drop in order to supply the current you are drawing. .........................................
That is a common misconception. It's the voltage and frequency at the input that determines whether the core saturates, not the load current. Since the load current generates flux in opposite direction to the input current, there is no net change in core flux due to load current. (Look at the transformer winding polarities versus current direction in each winding to see this.) The main thing that limits transformer current is the winding resistance and the heat build-up from the IR losses.
 
Since the load current generates flux in opposite direction to the input current, there is no net change in core flux due to load current.

That was the heart of my question...What happens if your secondary coil has more turns and a lower resistance do to using a better conductor material than used in the primary coil? Is the flux in the core now unbalanced, thus eliminated your advantage?
 
Njguy
I think that you are a bit confused, the flux in the core is proportional to Ampere.Turns.

So 1 amp flowing through 100 turns of thin very resitive wire, will produce exactly the same magnetic flux as 1 amp flowing through 100 turns of superconducting unobtainium.

The resistive wire will dissipate a lot of heat, where the superconductor will dissipate none (in a theoretical world), but the reflected load from the secondary of the transformer back to the primary will be the same.

JimB
 
Njguy
I think that you are a bit confused, the flux in the core is proportional to Ampere.Turns.

So 1 amp flowing through 100 turns of thin very resitive wire, will produce exactly the same magnetic flux as 1 amp flowing through 100 turns of superconducting unobtainium.

The resistive wire will dissipate a lot of heat, where the superconductor will dissipate none (in a theoretical world), but the reflected load from the secondary of the transformer back to the primary will be the same.

JimB
You not at all reading what I am saying. I am saying MORE turns in the secondary coil with LESS resistance. = More turns with more current. Obviously you can't get free energy, so where would the loss come from? I am assuming from the flux being unbalanced in favor of the secondary coil overpowering the primary and eliminating your gains.
 
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You not at all reading what I am saying.
I have read it, but obviously don't understand it. It is obviously me who is a bit confused.

JimB
 
You not at all reading what I am saying. I am saying MORE turns in the secondary coil with LESS resistance. = More turns with more current. Obviously you can't get free energy, so where would the loss come from? I am assuming from the flux being unbalanced in favor of the secondary coil overpowering the primary and eliminating your gains.

Hi,

Let me run this by you. You make a transformer with the following characteristics using ordinary copper wire:
Primary wire diameter 0.02 inches
Secondary wire diameter 0.06 inches
Primary wire turns:100
Secondary wire turns: 200

So we have a step up transformer that someone who did not know how to design transformers created. The secondary turns are greater than the primary turns so it's a step up, yet the secondary wire diameter means the secondary wire can handle more current. So that doesnt make sense either because the secondary will have 1/2 the current of the primary current, so we dont need a wire diameter that is 3 times higher, we can actually get away with a wire diameter that is less than the primary because of that current fact.
The bottom line here is that the secondary current Is will be Ip/N where N is the turns ratio from primary to secondary, so for a step up as above we'd have N=2 so the secondary current would be Is=Ip/2 no matter how good the wire was that we used.

If we did design a transformer like that with primary wire 0.02 inches and 100 turns and we had wire that was 3 times less resistive (the unobtainium here) on the secondary then we would use a diameter that is LESS than the primary because it only has to handle 1/2 the primary current.

Another example, we design a transformer with 100 turns primary and 100 turns secondary. We use copper on the primary, we use silver on the secondary. We've wasted the silver wire because the current in the secondary will be the same as the primary.

Let's do a few calculations.
Here, r is resistivity and RT is total resisance and V1 is primary wire voltage drop with load and V2 is secondary voltage drop with load.

r1=6.8e-7 primary resistivity in Ohm inches
r2=3.4e-7 secondary resistivity in Ohm inches
D1=0.02 primary wire diameter in inches
D2=0.02 secondary wire diameter in inches
L1=120 length of primary winding wire in inches
L2=120*2 length of secondary winding wire in inches
RT1=4*r1*L1/(D1^2*pi) total resistance of primary wire in Ohms
RT2=4*r2*L2/(D2^2*pi) total resistance of secondary wire in Ohms
Ip=1 primary current in Amperes
Is=1/2 secondary current in Amperes
V1=Ip*RT1 primary wire voltage drop with load in volts
V2=Is*RT2 secondary wire voltage drop with load in volts
P1=V1*Ip primary wire power loss in watts
P2=V2*Is secondary wire power loss in watts
Doing these calculations we get:
P1=0.25974 watts
P2=0.064935 watts

So we see that the secondary has much less power being lost in it, so this would not be as good of a design as it could be because we are using a wire diameter that is too large for the secondary, and thus wasting the unobtainium wire material.

Knowing this maybe we use trial and error to get to the right value. To start with, we decrease the wire diameter of the secondary by a factor of 1/2, making it now 0.01 inches in diameter. Now we do the calculations over again and we get:
P1=0.25974
P2=0.25974

so now we loose equal power in both windings. In real life we might increase the primary wire diameter slightly because we now have more room in the core window area, so we make the transformer more efficient and also get a higher power rating for it that way.

The main point however is still the same, we calculate the secondary current by using:
Is=Ip/N

and that has nothing to do with the wire itself, for the most part.

Maybe you have some specific reason for asking this question, as to why you think the secondary should somehow violate nature. It's just fine to have curiosity like that but you should explain why you think this in more specific terms so we can address those issues directly. Perhaps you've discovered something new, and in that case we'd want to know what it is.
 
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You not at all reading what I am saying. I am saying MORE turns in the secondary coil with LESS resistance. = More turns with more current. Obviously you can't get free energy, so where would the loss come from? I am assuming from the flux being unbalanced in favor of the secondary coil overpowering the primary and eliminating your gains.
The flux is never unbalanced. The current flowing in the primary always is just sufficient to generate a flux that counteracts the flux cause by the current in the secondary. If you change the number of secondary turns, or the secondary resistance, or the secondary load, then the secondary current will change, but then the primary current changes in response so that the two fluxes are still equal and opposite.

You are trying to imagine a situation that cannot occur. Remember that the secondary current is determined by the secondary open-circuit voltage and the sum of secondary resistance and the load resistance (as well as the reflected primary resistance).
 
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